Let $D$ denote a closed two dimensional figure as: $D=2+iT\to 2\to 2-\delta\to 2-\delta+i(T-\delta)\to \frac{1}{2}+\epsilon+i(T-\delta)\to\frac{1}{2}+\epsilon\to\frac{1}{2}-\epsilon\to \frac{1}{2}-\epsilon+iT\to 2+iT$ where $0<\epsilon<\frac{1}{2}$ and $T>3$. Assume that the Riemann zeta function, $\zeta(s)$ is non zero on the boundary of $D$ and there exists $\delta_0>0$ such that $\zeta(s)$ is non zero on and inside the $L-$ shaped region $L_\delta$, where $L_\delta=2+iT\to 2\to 2-\delta\to 2-\delta+i(T-\delta)\to \frac{1}{2}-\epsilon+i(T-\delta)\to\frac{1}{2}-\epsilon+iT\to 2+iT$ whenever $0<\delta<\delta_0$.
I need to apply Littlewood's lemma for $\zeta(s)$ in $D$.
Question: How do we define logarithm for all points $s$ in $D$ so that we can apply Littlewood's lemma for $\zeta(s)$ in $D$. Note that $D$ is not a rectangle so normal definition of logarithm as in the usual Littlewood's lemma might not work.
I tried the following: version of Littlewood's Lemma for $D$: Let $D$ be defined as above and $\zeta(s)$ be a holomorphic function in $D$ not vanishing on the boundary of $D$. Then \begin{equation}\sum_{\rho\in D} \text{dist}(\rho)=\frac{1}{2\pi i} \int_{\partial{D}}\log \zeta (s)\ ds \tag{1}\end{equation} where in the above integral we go through $\partial{D}$ in negative direction (clockwise), $\rho$ are the zeros of $\zeta(s)$ inside $D$ counted with multiplicity, $\text{dist}(\rho)$ denotes the distance of $\rho$ to the left side of $D$ and $\log \zeta(s)$ is defined by continuous variation as follows: We start with a particular determination on $\sigma=2$, and obtain the value at other points of $D$ by continuous variation from $\log \zeta(2+it)$ down to $\log \zeta(2)$ and then left to (a point in between $2$ and $2-\delta$) $\log \zeta(2a+(2-\delta)(1-a))$ (where $0\leq a\leq 1$) then upwards to $\log \zeta((2a+(2-\delta)(1-a))+it')$ (where $t'=(T-\delta)b+T(1-b)$ where $0\leq b\leq 1$) then left to $\log \zeta\left(\left(\frac{1}{2}-\epsilon\right)c+\left(\frac{1}{2}+\epsilon\right)(1-c)+it'\right)$ (where $0\leq c\leq 1$) and then down to $\log \zeta(\sigma+it)$ till we reach $\sigma+it$. Mathematically we have for $s\in D$ and $A=2a+(2-\delta)(1-a)$ , $B=\left(\frac{1}{2}-\epsilon\right)c+\left(\frac{1}{2}+\epsilon\right)(1-c)$ \begin{align}\log \zeta(\sigma+it)&=\log \zeta(2+it)\notag\\& +\left(\int_{2+it}^{2}+\int_{2}^{A}+\int_{A}^{A+it'}+\int_{A+it'}^{B+it'}+\int_{B+it'}^{\sigma+it}\right) \frac{\zeta'(s)}{\zeta(s)}\ ds \tag{2}\end{align} provided that $t$ and $t'$ are not the ordinate of zeros. If, however, this path would cross a zero or pole of $\zeta(s)$, we take $\log \zeta(s)$ to be $\log \zeta(s\pm i0)$ according as we approach the path from above or below.
Does the definition of logarithm needs any changes?
Thank you.